Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $y = \dfrac{z + 7}{z^2 + 4z} \times \dfrac{z^3 + 15z^2 + 50z}{5z^2 + 85z + 350} $
First factor out any common factors. $y = \dfrac{z + 7}{z(z + 4)} \times \dfrac{z(z^2 + 15z + 50)}{5(z^2 + 17z + 70)} $ Then factor the quadratic expressions. $y = \dfrac {z + 7} {z(z + 4)} \times \dfrac {z(z + 10)(z + 5)} {5(z + 10)(z + 7)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac {(z + 7) \times z(z + 10)(z + 5) } {z(z + 4) \times 5(z + 10)(z + 7) } $ $y = \dfrac {z(z + 10)(z + 5)(z + 7)} {5z(z + 10)(z + 7)(z + 4)} $ Notice that $(z + 10)$ and $(z + 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac {z\cancel{(z + 10)}(z + 5)(z + 7)} {5z\cancel{(z + 10)}(z + 7)(z + 4)} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $y = \dfrac {z\cancel{(z + 10)}(z + 5)\cancel{(z + 7)}} {5z\cancel{(z + 10)}\cancel{(z + 7)}(z + 4)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $y = \dfrac {z(z + 5)} {5z(z + 4)} $ $ y = \dfrac{z + 5}{5(z + 4)}; z \neq -10; z \neq -7 $